Question

Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is ______.

पर्याय

• `(4πd)/λ (1 - 1/n^2 sin^2 θ)^(1/2) + π`

• `(4πd)/λ (1 - 1/n^2 sin^2 θ)^(1/2)`

• `(4πd)/λ (1 - 1/n^2 sin^2 θ)^(1/2) + π/2`

• `(4πd)/λ (1 - 1/n^2 sin^2 θ)^(1/2) + 2π`

Answer 1

Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is `underline((4πd)/λ (1 - 1/n^2 sin^2 θ)^(1/2) + π)`.

Explanation:

If slab of a glass is placed in air, the wave reflected from the upper surface (from a denser medium) sutlers a sudden phase change of π, while the wave reflected from the lower surface (from a rarer medium) suffers no such phase change.

It is useful to draw an analogy between reflected light waves and the reflections of a transverse wave on a stretched string when the wave meets a boundary:

Figure (a) shows that ray reflecting from a medium of higher refractive index undergoes a 180° phase change.

Now consider the diagram, the ray (P) is incident at an angle θ and gets reflected in the direction P' and refracted in the direction P". Due to reflection from the glass medium, there is a phase change of π.

Time taken to travel along OP"

Δt = `(OP^")/v = (d/(cos  r))/(c/n) = (nd)/(c cos r)`

From Snell's law, n = `sin θ/sin r`

⇒ `sin r = sin θ/n`

`cos r = sqrt(1 - sin^2r) = sqrt(1 - (sin^2θ)/v^2)`

Phase difference, `Δphi = (2pi)/T xx Δt` ⇒ `Δphi = (2pind)/λ (1 - (sin^2θ)/n^2)^(-1/2)`

So, net phase difference = `Δphi + pi`

⇒ `Δphi_("net") = (4pid)/λ (1 - 1/n^2 sin^2θ) + pi`